A301336 -id:A301336 - cp's OEIS frontend

A301896 a(n) = product of total number of 0's and total number of 1's in binary expansions of 0, ..., n.

0, 1, 4, 8, 20, 35, 54, 72, 117, 165, 221, 280, 352, 425, 504, 576, 726, 875, 1036, 1200, 1386, 1575, 1776, 1976, 2214, 2451, 2700, 2944, 3216, 3479, 3750, 4000, 4455, 4897, 5355, 5808, 6300, 6789, 7296, 7800, 8364, 8925, 9504, 10080, 10695, 11305, 11931, 12544, 13260, 13965, 14688

Offset: 0

Ilya Gutkovskiy, Mar 28 2018

			+---+-----+---+---+---+---+----------+
| n | bin.|0's|sum|1's|sum|   a(n)   |
+---+-----+---+---+---+---+----------+
| 0 |   0 | 1 | 1 | 0 | 0 | 1*0 =  0 |
| 1 |   1 | 0 | 1 | 1 | 1 | 1*1 =  1 |
| 2 |  10 | 1 | 2 | 1 | 2 | 2*2 =  4 |
| 3 |  11 | 0 | 2 | 2 | 4 | 2*4 =  8 |
| 4 | 100 | 2 | 4 | 1 | 5 | 4*5 = 20 |
| 5 | 101 | 1 | 5 | 2 | 7 | 5*7 = 35 |
| 6 | 110 | 1 | 6 | 2 | 9 | 6*9 = 54 |
+---+-----+---+---+---+---+----------+
bin. - n written in base 2;
0's - number of 0's in binary expansion of n;
1's - number of 1's in binary expansion of n;
sum - total number of 0's (or 1's) in binary expansions of 0, ..., n.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A000120, A000788, A023416, A059015, A071295, A301336.

b:= proc(n) option remember; `if`(n=0, [1, 0], b(n-1)+
     (l-> [add(1-i, i=l), add(i, i=l)])(Bits[Split](n)))
    end:
a:= n-> (l-> l[1]*l[2])(b(n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Mar 01 2023

Accumulate[DigitCount[Range[0, 50], 2, 0]] Accumulate[DigitCount[Range[0, 50], 2, 1]]

def A301896(n): return (2+(n+1)*(m:=(n+1).bit_length())-(1<Chai Wah Wu, Mar 01 2023

def A301896(n): return (a:=(n+1)*n.bit_count()+(sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1))*(2+(n+1)*(t:=(n+1).bit_length())-(1<Chai Wah Wu, Nov 11 2024

a(n) = A059015(n)*A000788(n).

a(2^k-1) = 2^(k-2)*(2^k*(k - 2) + 4)*k.

A334841 a(0) = 0; for n > 0, a(n) = (number of 1's and 3's in base 4 representation of n) - (number of 0's and 2's in base 4 representation of n).

Original entry on oeis.org

0, 1, -1, 1, 0, 2, 0, 2, -2, 0, -2, 0, 0, 2, 0, 2, -1, 1, -1, 1, 1, 3, 1, 3, -1, 1, -1, 1, 1, 3, 1, 3, -3, -1, -3, -1, -1, 1, -1, 1, -3, -1, -3, -1, -1, 1, -1, 1, -1, 1, -1, 1, 1, 3, 1, 3, -1, 1, -1, 1, 1, 3, 1, 3, -2, 0, -2, 0, 0, 2, 0, 2, -2, 0, -2, 0, 0, 2, 0, 2, 0, 2, 0, 2, 2, 4, 2, 4, 0

Offset: 0

Alexander Van Plantinga, May 13 2020

Values are even for base 4 representations of n with an even number of digits, and odd for base 4 representations of n with an odd number of digits, except for a(0).

			      n in    #odd    #even
  n  base 4  digits - digits = a(n)
  =  ======  =======================
  0    0        0   -            0
  1    1        1   -    0   =   1
  2    2        0   -    1   =  -1
  3    3        1   -    0   =   1
  4   10        1   -    1   =   0
  5   11        2   -    0   =   2
  6   12        1   -    1   =   0
  7   13        2   -    0   =   2

Cf. A053737, A010065, A037863, A007090, A301336, A333596.

a:= n-> `if`(n=0, 0, add(`if`(i in [1, 3], 1, -1), i=convert(n, base, 4))):
seq(a(n), n=0..100);  # Alois P. Heinz, May 30 2020

a[0] = 0; a[n_] := Total[-(-1)^(r = Range[0, 3]) * DigitCount[n, 4, r]]; Array[a, 100, 0] (* Amiram Eldar, May 13 2020 *)
Join[{0},Table[Total[If[EvenQ[#],-1,1]&/@IntegerDigits[n,4]],{n,90}]] (* Harvey P. Dale, Sep 06 2020 *)

a(n) = my(ret=0); if(n,forstep(i=0,logint(n,2),2, if(bittest(n,i),ret++,ret--))); ret; \\ Kevin Ryde, May 24 2020

import numpy as np
def qnary(n):
    e = n//4
    q = n%4
    if n == 0 : return 0
    if e == 0 : return q
    if e != 0 : return np.append(qnary(e), q)
m = 400
v = [0]
for i in range(1, m+1) :
    t = np.array(qnary(i))
    t[t%2 != 0] = 1
    t[t%2 == 0] = -1
    v = np.append(v, np.sum(t))

def A334841(n):
    return 2*bin(n)[-1:1:-2].count('1')-(len(bin(n))-1)//2 if n > 0 else 0 # Chai Wah Wu, Sep 03 2020

qnary = function(n, e, q){
  e = floor(n/4)
  q = n%%4
  if(n == 0 ){return(0)}
  if(e == 0){return(q)}
  else{return(c(qnary(e), (q)))}
}
m = 400
s = seq(2, m)
v = c(0)
for(i in s){
  x = qnary(i-1)
  x[which(x%%2!=0)] = 1
  x[which(x%%2==0)] = -1
  v[i] = sum(x)
}

a(n) = 2*A139351(n) - A110591(n), n>0. - R. J. Mathar, Sep 02 2020

A333596 a(0) = 0; for n > 0, a(n) = a(n-1) + (number of 1's and 3's in base-4 representation of n) - (number of 0's and 2's in base-4 representation of n).

Original entry on oeis.org

0, 1, 0, 1, 1, 3, 3, 5, 3, 3, 1, 1, 1, 3, 3, 5, 4, 5, 4, 5, 6, 9, 10, 13, 12, 13, 12, 13, 14, 17, 18, 21, 18, 17, 14, 13, 12, 13, 12, 13, 10, 9, 6, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 9, 10, 13, 12, 13, 12, 13, 14, 17, 18, 21, 19, 19, 17, 17, 17, 19, 19, 21, 19, 19

Offset: 0

Alexander Van Plantinga, Mar 27 2020

Local maxima values minus 1 are divisible by 4.

For a digit-wise recurrence, it's convenient to sum n terms so b(n) = a(n-1) = Sum_{i=0..n-1} A334841(i). Then b(4n+r) = 4*b(n) + r*A334841(n) + (1 if r even), for 0 <= r <= 3 and 4n+r >= 1. This is 4 copies of terms 0..n-1 and r copies of the following n. The new lowest digits cancel when r is odd, or net +1 when r is even. Repeatedly expanding gives the PARI code below. - Kevin Ryde, Jun 02 2020

			      n in    #odd    #even
  n  base 4  digits - digits + a(n-1) = a(n)
  =  ======  ===============================
  0    0        0   -                     0
  1    1        1   -    0   +    0   =   1
  2    2        0   -    1   +    1   =   0
  3    3        1   -    0   +    0   =   1
  4   10        1   -    1   +    1   =   1
  5   11        2   -    0   +    1   =   3
  6   12        1   -    1   +    3   =   3
  7   13        2   -    0   +    3   =   5

Cf. A053737, A010065, A037863, A268289, A007090, A231664, A301336, A334841.

a:= proc(n) option remember; `if`(n=0, 0, a(n-1) +add(
     `if`(i in [1, 3], 1, -1), i=convert(n, base, 4)))
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, May 30 2020

f[n_] := Total[(-1)^(r = Range[0, 3]) * DigitCount[n, 4, r]]; a[0] = 0; a[n_] := a[n] = a[n - 1] - f[n]; Array[a, 100, 0] (* Amiram Eldar, Apr 24 2020 *)

a(n) = my(v=digits(n+1,4),s=0); for(i=1,#v, my(t=v[i]); v[i]=t*s+!(t%2); s-=(-1)^t); fromdigits(v,4); \\ Kevin Ryde, May 30 2020

b(n)=my(d=digits(n,4)); -sum(i=1,#d,(-1)^d[i])
first(n)=my(s); concat(0,vector(n,k,s+=b(k))) \\ Charles R Greathouse IV, Jul 04 2020

import numpy as np
def qnary(n):
    e = n//4
    q = n%4
    if n == 0 : return 0
    if e == 0 : return q
    if e != 0 : return np.append(qnary(e), q)
m = 400
v = [0]
for i in range(1,m+1) :
    t = np.array(qnary(i))
    t[t%2 != 0] = 1
    t[t%2 == 0] = -1
    v = np.append(v, np.sum([np.sum(t), v[i-1]]))

from itertools import accumulate
def A334841(n):
    return 2*bin(n)[-1:1:-2].count('1')-(len(bin(n))-1)//2 if n > 0 else 0
A333596_list = list(accumulate(A334841(n) for n in range(10000))) # Chai Wah Wu, Sep 03 2020

qnary = function(n, e, q){
  e = floor(n/4)
  q = n%%4
  if(n == 0 ){return(0)}
  if(e == 0){return(q)}
  else{return(c(qnary(e), (q)))}
}
m = 400
s = seq(2,m)
v = c(0)
for(i in s){
  x = qnary(i-1)
  x[which(x%%2!=0)] = 1
  x[which(x%%2==0)] = -1
  v[i] = sum(x,v[i-1])
}

cp's OEIS Frontend

A301896 a(n) = product of total number of 0's and total number of 1's in binary expansions of 0, ..., n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Python

Formula

A334841 a(0) = 0; for n > 0, a(n) = (number of 1's and 3's in base 4 representation of n) - (number of 0's and 2's in base 4 representation of n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Python

R

Formula

A333596 a(0) = 0; for n > 0, a(n) = a(n-1) + (number of 1's and 3's in base-4 representation of n) - (number of 0's and 2's in base-4 representation of n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Python

Python

R